/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 75 Some straightforward but slightl... [FREE SOLUTION] | 91Ó°ÊÓ

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Some straightforward but slightly tedious algebra shows that $$ \text { SSResid }=\left(1-r^{2}\right) \sum(y-\bar{y})^{2} $$ from which it follows that $$ s_{e}=\sqrt{\frac{n-1}{n-2}}\left(\sqrt{1-r^{2}}\right) s_{y} $$ Unless \(n\) is quite small, \((n-1) /(n-2) \approx 1\), so $$ s_{e} \approx\left(\sqrt{1-r^{2}}\right) s_{y} $$ a. For what value of \(r\) is \(s_{e}\) as large as \(s_{y}\) ? What is the equation of the least-squares line in this case? b. For what values of \(r\) will \(s_{e}\) be much smaller than \(s_{y}\) ?

Short Answer

Expert verified
For \(s_{e}\) to be as large as \(s_{y}\), \(r\) must be 0 and the least-squares line is \(y = \bar{y}\). For \(s_{e}\) to be much smaller than \(s_{y}\), \(r\) must be close to -1 or 1.

Step by step solution

01

Part a: Determine \(r\) when \(s_{e}=s_{y}\)

Setting \(s_{e}=s_{y}\) in the approximation formula gives \[ s_{e} = s_{y} \implies \sqrt{1-r^{2}}s_{y} = s_{y} \] Simplifying this results in \(\sqrt{1-r^{2}} = 1\). Squaring both sides to eliminate the square root yields \(1-r^{2}= 1\). Solving for \(r\), we find that \(r=0\).
02

Part a: Least-Squares Line Equation when \(r=0\)

When \(r=0\), the equation of the least-squares line or regression line is \(y = \bar{y}\), where \(\bar{y}\) is the mean of the dependent variable \(y\). There is no correlation between the dependent and independent variables, and the slope of the line is zero.
03

Part b: Determine \(r\) when \(s_{e}

When \(s_{e} << s_{y}\), the fraction \(\sqrt{1-r^{2}}\) is much smaller than 1, which implies that \(1-r^{2}\) is much smaller than 1. This occurs when \(r\) is either close to 1 or -1, that is, when the correlation between the dependent and independent variables is very strong either positively or negatively.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sum of Squares Residual
The concept of the sum of squares residual (SSResid) is critical in regression analysis. It helps us understand the variation in the dependent variable that is not explained by the regression model. Mathematically, SSResid is calculated by taking the difference between the actual and the predicted values, then squaring each of these differences, and finally summing them all up. In essence, it measures the discrepancy between the data and the estimation model.

Visually, consider each data point on a scatter plot; the vertical distance from this point to the line of best fit (the least squares line) is its individual 'residual'. When we talk about 'sum of squares', we're referring to the sum of each of these squared residuals. This is important because in regression, our goal is typically to minimize these residuals, thus minimizing the SSResid, to get the most accurate estimation line possible.

As we get a better fitted line, the SSResid will decrease which indicates a more reliable model with less unexplained variance. Consequently, our model's predictions become more trustworthy.
Standard Error of the Estimate
The standard error of the estimate, denoted as 'se', is a measure of the accuracy of predictions made with a regression line. Essentially, it's the average distance that the observed values fall from the regression line. Think of it as a ruler telling us how much error we can expect from our model when making predictions.

The formula given in the exercise \(s_{e} = \sqrt{1-r^{2}}s_{y}\) shows that the standard error of the estimate is proportional to the standard deviation of the dependent variable \(s_{y}\) and inversely proportional to the strength of the correlation coefficient \(r\). When \(r\) is low, indicating a weak correlation between variables, \(s_{e}\) is closer to \(s_{y}\), signaling higher error margins and less precise predictions. In contrast, a high \(r\) value suggests \(s_{e}\) is smaller than \(s_{y}\), indicating a stronger relationship and more confidence in the predictions.
Least Squares Line
The least squares line is the foundation of linear regression analysis. It's the straight line that best fits the data points on a scatter plot, chosen such that it minimizes the sum of the squares of the residuals (the distances between the line and the observed data points). To find this line, we use the least squares method, a form of mathematical optimization.

When the correlation coefficient \(r\) is zero, as discussed in the exercise, our least squares line is a horizontal line at the mean of all Y-values, indicating no relationship between the variables. The slope is zero, so for every X, the best prediction we can make for Y is simply the average of Y. As the value of \(r\) deviates from zero, reaching towards -1 or 1, our least squares line tips and tilts, indicating a negative or positive relationship between the variables, respectively. The slope of the line reflects the strength and direction of this relationship.
Correlation Coefficient
The correlation coefficient \(r\) is a statistical measure that calculates the strength and direction of a linear relationship between two variables. Values range from -1 to 1, where 1 means a perfect positive linear correlation, -1 indicates a perfect negative linear correlation, and 0 implies no linear correlation.

When we perform regression analysis, \(r\) plays a central role. A high absolute value of \(r\), close to 1 or -1, suggests that the regression line provides a good fit to the data, and therefore, we can make fairly accurate predictions. On the flip side, an \(r\) value near 0 means our regression model does not explain the variability of the data well. We use the square of this coefficient, known as coefficient of determination (\(r^2\)), to represent the proportion of the variance in the dependent variable that is predictable from the independent variable.

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Most popular questions from this chapter

The article "Improving Fermentation Productivity with Reverse Osmosis" (Food Technology [1984]: \(92-96\) ) gave the following data (read from a scatterplot) on \(y=\) glucose concentration \((\mathrm{g} / \mathrm{L})\) and \(x=\) fermentation time (days) for a blend of malt liquor. $$ \begin{array}{lrrrrrrrr} x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ y & 74 & 54 & 52 & 51 & 52 & 53 & 58 & 71 \end{array} $$ a. Use the data to calculate the estimated regression line. b. Do the data indicate a linear relationship between \(y\) and \(x\) ? Test using a \(.10\) significance level. c. Using the estimated regression line of Part (a), compute the residuals and construct a plot of the residuals versus \(x\) (that is, of the \((x\), residual \()\) pairs. d. Based on the plot in Part (c), do you think that a linear model is appropriate for describing the relationship between \(y\) and \(x\) ? Explain.

The article "Effects of Enhanced UV-B Radiation on Ribulose-1,5-Biphosphate, Carboxylase in Pea and Soybean" (Environmental and Experimental Botany [1984]: 131-143) included the accompanying data on pea plants, with \(y=\) sunburn index and \(x=\) distance \((\mathrm{cm})\) from an ultraviolet light source. \(\begin{array}{lllllllll}x & 18 & 21 & 25 & 26 & 30 & 32 & 36 & 40 \\ y & 4.0 & 3.7 & 3.0 & 2.9 & 2.6 & 2.5 & 2.2 & 2.0 \\ x & 40 & 50 & 51 & 54 & 61 & 62 & 63 & \\ y & 2.1 & 1.5 & 1.5 & 1.5 & 1.3 & 1.2 & 1.1 & \end{array}\) $$ \begin{array}{lc} \sum x=609 & \sum y=33.1 \quad \sum x^{2}=28,037 \\ \sum y^{2}=84.45 & \sum x y=1156.8 \end{array} $$ Estimate the mean change in the sunburn index associated with an increase of \(1 \mathrm{~cm}\) in distance in a way that includes information about the precision of estimation.

The accompanying data on \(x=\) U.S. population (millions) and \(y=\) crime index (millions) appeared in the article "The Normal Distribution of Crime" (Journal of Police Science and Administration \([1975]: 312-318)\). The author comments that "The simple linear regression analysis remains one of the most useful tools for crime prediction." When observations are made sequentially in time, the residuals or standardized residuals should be plotted in time order (that is, first the one for time \(t=1\) ( 1963 here), then the one for time \(t=2\), and so on ). Notice that here \(x\) increases with time, so an equivalent plot is of residuals or standardized residuals versus \(x\). Using \(\hat{y}=47.26+.260 x\), calculate the residuals and plot the \((x\), residual) pairs. Does the plot exhibit a pattern that casts doubt on the appropriateness of the simple linear regression model? Explain. \(\begin{array}{lrrrrrr}\text { Year } & 1963 & 1964 & 1965 & 1966 & 1967 & 1968 \\ x & 188.5 & 191.3 & 193.8 & 195.9 & 197.9 & 199.9 \\ y & 2.26 & 2.60 & 2.78 & 3.24 & 3.80 & 4.47 \\\ \text { Year } & 1969 & 1970 & 1971 & 1972 & 1973 & \\ x & 201.9 & 203.2 & 206.3 & 208.2 & 209.9 & \\ y & 4.99 & 5.57 & 6.00 & 5.89 & 8.64 & \end{array}\)

The accompanying data on \(x=\) treadmill run time to exhaustion (min) and \(y=20-\mathrm{km}\) ski time (min) were taken from the article "Physiological Characteristics and Performance of Top U.S. Biathletes" (Medicine and \(S c i\) ence in Sports and Exercise \([1995]: 1302-1310):\) \(\begin{array}{lrrrrrr}x & 7.7 & 8.4 & 8.7 & 9.0 & 9.6 & 9.6 \\ y & 71.0 & 71.4 & 65.0 & 68.7 & 64.4 & 69.4 \\ x & 10.0 & 10.2 & 10.4 & 11.0 & 11.7 & \\ y & 63.0 & 64.6 & 66.9 & 62.6 & 61.7 & \\ \sum x & =106.3 & \sum x^{2}=1040.95 & & \\ \sum y= & 728.70 & \sum x y=7009.91 & \sum y^{2}=48390.79\end{array}\) a. Does a scatterplot suggest that the simple linear regression model is appropriate? b. Determine the equation of the estimated regression line, and draw the line on your scatterplot. c. What is your estimate of the average change in ski time associated with a 1 -min increase in treadmill time? d. What would you predict ski time to be for an individual whose treadmill time is \(10 \mathrm{~min} ?\) e. Should the model be used as a basis for predicting ski time when treadmill time is \(15 \mathrm{~min}\) ? Explain. f. Calculate and interpret the value of \(r^{2}\). g. Calculate and interpret the value of \(s_{e}\)

Exercise \(13.8\) gave data on \(x=\) treadmill run time to exhaustion and \(y=20-\mathrm{km}\) ski time for a sample of 11 biathletes. Use the accompanying MINITAB output to answer the following questions. The regression equation is ski \(=-88.8-2.33\) tread \(\begin{array}{lrrrr}\text { Predictor } & \text { Coef } & \text { Stdev } & \text { t-ratio } & p \\\ \text { Constant } & 88.796 & 5.750 & 15.44 & 0.000 \\ \text { tread } & 2.3335 & 0.5911 & 3.95 & 0.003 \\ s=2.188 & \text { R-sq }=63.4 \% & \text { R-sq }(a d j)=59.3 \% & \end{array}\) Analysis of Variance \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { p } \\ \text { Regression } & 1 & 74.630 & 74.630 & 15.58 & 0.003 \\ \text { Error } & 9 & 43.097 & 4.789 & & \\ \text { Total } & 10 & 117.727 & & & \end{array}\) a. Carry out a test at significance level \(.01\) to decide whether the simple linear regression model is useful. b. Estimate the average change in ski time associated with a 1 -minute increase in treadmill time, and do so in a way that conveys information about the precision of estimation. c. MINITAB reported that \(s_{a+b(10)}=.689\). Predict ski time for a single biathlete whose treadmill time is \(10 \mathrm{~min}\), and do so in a way that conveys information about the precision of prediction. d. MINITAB also reported that \(s_{a+b(11)}=1.029 .\) Why is this larger than \(s_{a+b(10) ?}\)

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